Engineering Materials vol 2 Part 4 doc

To summar- ise, if random fluctuations in the liquid give crystals with r > r* then stable nuclei will form, and solidification can begin.. As we said earlier, small groups of liquid ato

from the solid–liquid interface (for the same sort of reason that your hands lose heatmuch more rapidly if you wear gloves than if you wear mittens) And the fastersolidification that we get as a consequence “pays for” the high boundary energy Largedriving forces produce fine dendrites – which explains why one can hardly see thedendrites in an iced lollipop grown in a freezer (−10°C); but they are obvious on afreezing pond (−1°C).

To summarise, the shapes of the grains and phases produced during tions reflect a balance between the need to minimise the total boundary energy and theneed to maximise the speed of transformation Close to equilibrium, when the drivingforce for the transformation is small, the grains and phases are primarily shaped bythe boundary energies Far from equilibrium, when the driving force for the transforma-

transforma-tion is large, the structure depends strongly on the mechanism of the transformatransforma-tion Further, even the scale of the structure depends on the driving force – the larger the

driving force the finer the structure

G A Chadwick, Metallography of Phase Transformations, Butterworth, 1972.

P G Shewmon, Diffusion in Solids, 2nd edition, TMS Publishers, 1989.

Problems

6.1 The solidification speed of salol is about 2.3 mm min–1 at 10°C Using eqn (6.15)

estimate the energy barrier q that must be crossed by molecules moving from

liquid sites to solid sites The melting point of salol is 43°C and its latent heat

of fusion is 3.2 × 10–20 J molecule–1 Assume that the molecular diameter is about

1 nm

Answer: 6.61 × 10–20 J, equivalent to 39.8 kJ mol–1

6.2 Glass ceramics are a new class of high-technology crystalline ceramic They aremade by taking complex amorphous glasses (like SiO2–Al2O3–Li2O) and makingthem devitrify (crystallise) For a particular glass it is found that: (a) no devitrificationoccurs above a temperature of 1000°C; (b) the rate of devitrification is a maximum

at 950°C; (c) the rate of devitrification is negligible below 700°C Give reasons forthis behaviour

6.3 Samples cut from a length of work-hardened mild steel bar were annealed forvarious times at three different temperatures The samples were then cooled toroom temperature and tested for hardness The results are given below

Annealing temperature Vickers hardness Time at annealing temperature

i.e the time for recrystal-lisation, tr, is given by

these crystals – or nuclei – form in the first place?

Nucleation in liquids

We begin by looking at how crystals nucleate in liquids Because of thermal agitationthe atoms in the liquid are in a state of continual movement From time to time a smallgroup of atoms will, purely by chance, come together to form a tiny crystal If the

liquid is above T m the crystal will, after a very short time, shake itself apart again But

if the liquid is below T m there is a chance that the crystal will be thermodynamicallystable and will continue to grow How do we calculate the probability of finding stable

nuclei below T m?

There are two work terms to consider when a nucleus forms from the liquid tions (6.1) and (6.2) show that work of the type ∆H (T m − T )/T m is available to helpthe nucleus form If ∆H is expressed as the latent heat given out when unit volume of

Equa-the solid forms, Equa-then Equa-the total available energy is (4/3)πr3

∆H (T m − T)/T m But this isoffset by the work 4πr2γSL needed to create the solid–liquid interface around the crys-tal The net work needed to form the crystal is then

W f= 4πr2γSL−

433

T m m

(7.1)

This result has been plotted out in Fig 7.1 It shows that there is a maximum value

for W f corresponding to a critical radius r* For r < r* (dW f /dr) is positive, whereas for

r > r* it is negative This means that if a random fluctuation produces a nucleus of size

r < r* it will be unstable: the system can do free work if the nucleus loses atoms and r decreases The opposite is true when a fluctuation gives a nucleus with r > r* Then, free work is done when the nucleus gains atoms, and it will tend to grow To summar- ise, if random fluctuations in the liquid give crystals with r > r* then stable nuclei will

form, and solidification can begin

To calculate r* we differentiate eqn (7.1) to give

Fig 7.1. The work needed to make a spherical nucleus.

for the critical radius

We are now in a position to go back and look at what is happening in the liquid inmore detail As we said earlier, small groups of liquid atoms are continually shakingthemselves together to make tiny crystals which, after a short life, shake themselvesapart again There is a high probability of finding small crystals, but a small probabil-ity of finding large crystals And the probability of finding crystals containing morethan 102 atoms (r ⱚ 1 nm) is negligible As Fig 7.2 shows, we can estimate the tem-

perature Thom at which nucleation will first occur by setting r* = 1 nm in eqn (7.3) Fortypical values of γSL, T m and ∆H we then find that T m − Thom≈ 100 K, so an enormousundercooling is needed to make nucleation happen

This sort of nucleation – where the only atoms involved are those of the material

itself – is called homogeneous nucleation It cannot be the way materials usually solidify

because (usually) an undercooling of 1°C or less is all that is needed Homogeneousnucleation has been observed in ultraclean laboratory samples But it is the exception,not the rule

Heterogeneous nucleation

Normally, when a pond of water freezes over, or when a metal casting starts to solidify,

nucleation occurs at a temperature only a few degrees below T How do we explain

Fig 7.2. Homogeneous nucleation will take place when the random crystals can grow, i.e when r > r*.

this easy nucleation? Well, liquids like pond water or foundry melts inevitably containsolid particles of dirt These act as catalysts for nucleation: they give a random crystal

a “foothold”, so to speak, and allow it to grow more easily It is this heterogeneous

nucleation which is responsible for solidification in all practical materials situations.Heterogeneous nucleation is most likely to occur when there is a strong tendencyfor the crystal to stick to the surface of the catalyst This sticking tendency can bedescribed by the angle of contact, θ, shown in Fig 7.3: the smaller θ, the better theadhesion Anyone who has tried to get electronic solder to stick to a strip of copperwill understand this well If the copper is tarnished the solder will just roll around as

a molten blob with θ = 180°, and will not stick to the surface at all If the tarnished

Fig 7.3. Heterogeneous nucleation takes place on the surface of a solid catalyst For the catalyst to be

effective there must be a strong tendency for it to be “wetted” by the crystal, i.e q must be small.

copper is fluxed then θ may decrease to 90°: the molten solder will stay put on thecopper but it will not spread Only when the copper is both clean and fluxed will θ bezero, allowing the solder to “wet” the copper.

If we know the contact angle we can work out r* quite easily We assume that the nucleus is a spherical cap of radius r and use standard mathematical formulae for the

area of the solid–liquid interface, the area of the catalyst–solid interface and the ume of the nucleus For 0 艋 θ 艋 90° these are:

12

Then, by analogy with eqn (7.1) we can write

W f= 2πr2(1 − cos θ)γSL+ πr2(1 − cos2θ)γCS− πr2(1 − cos2θ)γCL

12

3

3

πr

H T T T

m m

Note that this equation has two energy terms that did not appear in eqn (7.1) Thefirst, πr2(1 − cos2 θ)γCS, is the energy needed to create the new interface between thecatalyst and the solid The second, –πr2(1 − cos2 θ)γCL, is the energy released becausethe area of the catalyst–liquid interface is smaller after nucleation than it was before

As it stands, eqn (7.7) contains too many unknowns But there is one additionalpiece of information that we can use The interfacial energies, γSL, γCS and γCL act assurface tensions in just the way that a soap film has both a surface energy and asurface tension This means that the mechanical equilibrium around the edge of thenucleus can be described by the triangle of forces

for the critical radius in heterogeneous nucleation

If we compare eqns (7.11) and (7.3) we see that the expressions for the critical radiusare identical for both homogeneous and heterogeneous nucleation But the expressions

for the volume of the critical nucleus are not For homogeneous nucleation the critical

volume is

whereas for heterogeneous nucleation it is

Vhet* = 2π( * ) { cos cos }rhet 31− 3 θ + 1 3θ . (7.13)The maximum statistical fluctuation of 102 atoms is the same in both homogeneousand heterogeneous nucleation If Ω is the volume occupied by one atom in the nucleusthen we can easily see that

If the nucleus wets the catalyst well, with θ = 10°, say, then eqn (7.15) tells us that

r*het= 18.1r*hom In other words, if we arrange our 102 atoms as a spherical cap on a goodcatalyst we get a much bigger crystal radius than if we arrange them as a sphere And,

as Fig 7.4 explains, this means that heterogeneous nucleation always “wins” overhomogeneous nucleation

It is easy to estimate the undercooling that we would need to get heterogeneousnucleation with a 10° contact angle From eqns (7.11) and (7.3) we have

Fig 7.4. Heterogeneous nucleation takes place at higher temperatures because the maximum random fluctuation of 10 2 atoms gives a bigger crystal radius if the atoms are arranged as a spherical cap.

Nucleation in solids

Nucleation in solids is very similar to nucleation in liquids Because solids usuallycontain high-energy defects (like dislocations, grain boundaries and surfaces) newphases usually nucleate heterogeneously; homogeneous nucleation, which occurs indefect-free regions, is rare Figure 7.5 summarises the various ways in which nucleationcan take place in a typical polycrystalline solid; and Problems 7.2 and 7.3 illustratehow nucleation theory can be applied to a solid-state situation

Summary

In this chapter we have shown that diffusive transformations can only take place if

nuclei of the new phase can form to begin with Nuclei form because random atomic

vibrations are continually making tiny crystals of the new phase; and if the ature is low enough these tiny crystals are thermodynamically stable and will grow In

temper-homogeneous nucleation the nuclei form as spheres within the bulk of the material In

Fig 7.5. Nucleation in solids Heterogeneous nucleation can take place at defects like dislocations, grain boundaries, interphase interfaces and free surfaces Homogeneous nucleation, in defect-free regions, is rare.

heterogeneous nucleation the nuclei form as spherical caps on defects like solid surfaces,

grain boundaries or dislocations Heterogeneous nucleation occurs much more easilythan homogeneous nucleation because the defects give the new crystal a good “foothold”.Homogeneous nucleation is rare because materials almost always contain defects

Postscript

Nucleation – of one sort or another – crops up almost everywhere During winterplants die and people get frostbitten because ice nucleates heterogeneously insidecells But many plants have adapted themselves to prevent heterogeneous nucleation;they can survive down to the homogeneous nucleation temperature of −40°C The

“vapour” trails left by jet aircraft consist of tiny droplets of water that have nucleatedand grown from the water vapour produced by combustion Sub-atomic particles can

be tracked during high-energy physics experiments by firing them through heated liquid in a “bubble chamber”: the particles trigger the nucleation of gas bubbleswhich show where the particles have been And the food industry is plagued bynucleation problems Sucrose (sugar) has a big molecule and it is difficult to get it

super-to crystallise from aqueous solutions That is fine if you want super-to make caramel –this clear, brown, tooth-breaking substance is just amorphous sucrose But the sugarrefiners have big problems making granulated sugar, and will go to great lengths toget adequate nucleation in their sugar solutions We give examples of how nucleationapplies specifically to materials in a set of case studies on phase transformations inChapter 9

Further reading

D A Porter and K E Easterling, Phase Transformations in Metals and Alloys, 2nd edition, Chapman

and Hall, 1992.

G J Davies, Solidification and Casting, Applied Science Publishers, 1973.

G A Chadwick, Metallography of Phase Transformations, Butterworth, 1972.

7.1 The temperature at which ice nuclei form homogeneously from under-cooled water

is –40°C Find r* given that γ = 25 mJ m–2, ∆ H = 335 kJ kg–1, and Tm = 273 K.Estimate the number of H2O molecules needed to make a critical-sized nucleus.Why do ponds freeze over when the temperature falls below 0°C by only a fewdegrees?

[The density of ice is 0.92 Mg m–3 The atomic weights of hydrogen and oxygenare 1.01 and 16.00 respectively.]

Answers: r*, 1.11 nm; 176 molecules.

7.2 An alloy is cooled from a temperature at which it has a single-phase structure (α)

to a temperature at which the equilibrium structure is two-phase (α + β) Duringcooling, small precipitates of the β phase nucleate heterogeneously at α grainboundaries The nuclei are lens-shaped as shown below

Show that the free work needed to produce a nucleus is given by

1

43

to make a critical-sized nucleus given the following data: ∆H = 3.48 kJ mol–1;atomic weight = 47.90; Te – T = 30 K; T e = 882°C; γ = 0.1 Jm–2; density of the c.p.h.phase = 4.5 Mg m–3; θ = 5°

GB

α

α β

Chapter 8

Kinetics of structural change:

III – displacive transformations

Introduction

So far we have only looked at transformations which take place by diffusion: the

so-called diffusive transformations But there is one very important class of transformation – the displacive transformation – which can occur without any diffusion at all.

The most important displacive transformation is the one that happens in carbonsteels If you take a piece of 0.8% carbon steel “off the shelf” and measure its mechan-ical properties you will find, roughly, the values of hardness, tensile strength andductility given in Table 8.1 But if you test a piece that has been heated to red heat andthen quenched into cold water, you will find a dramatic increase in hardness (4 times

or more), and a big decrease in ductility (it is practically zero) (Table 8.1)

The two samples have such divergent mechanical properties because they haveradically different structures: the structure of the as-received steel is shaped by adiffusive transformation, but the structure of the quenched steel is shaped by a displacivechange But what are displacive changes? And why do they take place?

In order to answer these questions as directly as possible we begin by looking at

diffusive and displacive transformations in pure iron (once we understand how pure

iron transforms we will have no problem in generalising to iron–carbon alloys) Now,

as we saw in Chapter 2, iron has different crystal structures at different temperatures.Below 914°C the stable structure is b.c.c., but above 914°C it is f.c.c If f.c.c iron iscooled below 914°C the structure becomes thermodynamically unstable, and it tries tochange back to b.c.c This f.c.c → b.c.c transformation usually takes place by a diffu-sive mechanism But in exceptional conditions it can occur by a displacive mechanisminstead To understand how iron can transform displacively we must first look at thedetails of how it transforms by diffusion

Table 8.1 Mechanical properties of 0.8% carbon steel

Fig 8.1. The diffusive f.c.c → b.c.c transformation in iron The vertical axis shows the speed of the b.c.c.– f.c.c interface at different temperatures Note that the transformation can take place extremely rapidly, making

it very difficult to measure the interface speeds The curve is therefore only semi-schematic.

The diffusive f.c.c → b.c.c transformation in pure iron

We saw in Chapter 6 that the speed of a diffusive transformation depends strongly ontemperature (see Fig 6.6) The diffusive f.c.c → b.c.c transformation in iron shows thesame dependence, with a maximum speed at perhaps 700°C (see Fig 8.1) Now wemust be careful not to jump to conclusions about Fig 8.1 This plots the speed of anindividual b.c.c.–f.c.c interface, measured in metres per second If we want to know

the overall rate of the transformation (the volume transformed per second) then we need to know the area of the b.c.c.–f.c.c interface as well.

The total area of b.c.c.–f.c.c interface is obviously related to the number of b.c.c.nuclei As Fig 8.2 shows, fewer nuclei mean a smaller interfacial area and a smallervolume transforming per second Indeed, if there are no nuclei at all, then the rate oftransformation is obviously zero The overall rate of transformation is thus givenapproximately by

We know that the interfacial speed varies with temperature; but would we expect thenumber of nuclei to depend on temperature as well?

The nucleation rate is, in fact, critically dependent on temperature, as Fig 8.3 shows

To see why, let us look at the heterogeneous nucleation of b.c.c crystals at grain aries We have already looked at grain boundary nucleation in Problems 7.2 and 7.3.Problem 7.2 showed that the critical radius for grain boundary nucleation is given by